Electric charges push and pull on each other across empty space. The field is a way of saying what a charge would feel at every point.
🧲 Coulomb’s law
Two point charges exert equal and opposite forces on each other. Like charges repel and unlike charges attract. The force follows an inverse-square law.
F = k q₁ q₂ / r² k = 9 × 10⁹ N·m²/C²
q₁ = 2 μC, q₂ = 3 μC, r = 0.3 m
F = 9×10⁹ · (2×10⁻⁶)(3×10⁻⁶) / 0.3²
= 0.054 / 0.09 = 0.6 N
Double the distance → force ÷ 4
🖼️ See it
Double the distance and the force falls to a quarter: an inverse-square law.
🌐 The electric field
The electric field E at a point is the force per unit positive charge: E = F / q. A point charge Q makes a field E = kQ / r² pointing away from positive charge. Field lines show direction and are closer together where the field is stronger.
Q = 4 μC, r = 3 m
E = 9×10⁹ · 4×10⁻⁶ / 3² = 4000 N/C
force on a 0.5 μC charge there:
F = qE = 0.5×10⁻⁶ · 4000 = 0.002 N
Field is a vector. To find the field from several charges, add the individual field vectors, paying attention to direction.
🧪 Worked examples: adding fields and balancing forces
Field from two charges. Fields are vectors, so add them with direction. Between opposite charges, both fields point the same way and add.
+4 μC and −4 μC are 2.0 m apart; find the field at the midpoint (1.0 m from each)
each charge: E = kQ ÷ r² = 9.0 × 10⁹ × 4.0 × 10⁻⁶ ÷ 1² = 36 000 N/C
both point from + towards −, so E = 36 000 + 36 000 = 72 000 N/C
Inverse-square scaling. If the force is 0.60 N at 0.30 m, then at 0.10 m (one third of the distance) it is 3² = 9 times bigger: 5.4 N.
A charged droplet in balance (Millikan’s idea). The upward electric force equals the weight.
droplet with charge 2e = 3.2 × 10⁻¹⁹ C in a field of 3.0 × 10⁵ V/m
q E = m g → m = q E ÷ g = 3.2 × 10⁻¹⁹ × 3.0 × 10⁵ ÷ 10
= 9.6 × 10⁻¹⁵ kg
Zero field point. Between two like charges, the field is zero closer to the smaller charge, where the two fields cancel.
🎯 Exam focus: field lines and gravitational analogies
Field lines start on positive charges and end on negative ones. They never cross, and the closer they are, the stronger the field. Equipotential lines cross field lines at right angles.
electric gravitational
force law F = kQq ÷ r² F = GMm ÷ r²
field E = kQ ÷ r² g = GM ÷ r²
can it repel? yes (like charges) no (always attractive)
uniform field between parallel plates near the Earth’s surface
Common mistakes. Adding fields as scalars. Forgetting that the force on a negative charge is opposite to the field. Using the separation in cm or μC without converting. Writing E = F ÷ q with the sign of q ignored. Saying that a charged object only attracts (a neutral object is attracted by induced charge).
Units. E can be written in N C⁻¹ or V m⁻¹; they are the same.
✏️ Quick check · 1 of 11
Like charges
Two positive charges placed near each other will:
Opposites attract, likes do not.
✏️ Quick check · 2 of 11
Field direction
The electric field direction at a point is the direction of the force on a:
Convention.
✏️ Quick check · 3 of 11
Inverse square
If the distance between two charges triples, the force becomes:
Square the factor.
✍️ Type your answer · 4 of 11
q₁ = 2 μC and q₂ = 3 μC are 0.3 m apart (k = 9 × 10⁹).
What is the force between them, in N?
Enter an exact answer, without rounding.
N
Use F = kq₁q₂/r².
✍️ Type your answer · 5 of 11
Two charges 1 m apart exert 8 N on each other.
What is the force when the distance is doubled to 2 m?
Enter an exact answer, without rounding.
N
Inverse-square law.
✍️ Type your answer · 6 of 11
A point charge Q = 4 μC, at a point 3 m away (k = 9 × 10⁹).
What is the field strength, in N/C?
Enter an exact answer, without rounding.
N/C
E = kQ/r².
✍️ Type your answer · 7 of 11
A charge of 0.5 μC sits in a field of 4000 N/C.
What is the force on it, in N?
Enter an exact answer, without rounding.
N
F = qE.
✍️ Type your answer · 8 of 11
Charges of +4.0 μC and −4.0 μC are 2.0 m apart in a vacuum. Use k = 9.0 × 10⁹ N m² C⁻².
What is the size of the field at the midpoint between them, in N/C?
Enter an exact answer, without rounding.
N/C
The two fields point the same way.
✍️ Type your answer · 9 of 11
Two small charged spheres at a distance of 0.30 m exert a force of 0.60 N on each other.
What force would they exert at a distance of 0.10 m, in newtons?
Enter an exact answer, without rounding.
N
Use the inverse-square law.
✍️ Type your answer · 10 of 11
A droplet of charge 3.2 × 10⁻¹⁹ C is held still in a field of 3.0 × 10⁵ V/m (g = 10 m/s²).
What is its mass, in units of 10⁻¹⁵ kg?
Enter an exact answer, without rounding.
×10⁻¹⁵ kg
Balance qE with mg.
✏️ Quick check · 11 of 11
Field lines
Which statement about electric field lines is correct?
Think about a positive test charge.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The SI unit of electric field strength is:
Question 2 of 5
Charge is conserved, which means:
Question 3 of 5
Where are electric field lines closest together?
Question 4 of 5
Inside a conductor in electrostatic equilibrium the electric field is:
Question 5 of 5
An object becomes positively charged by:
Lesson 2 · Electricity
🔋 Electric Potential & Capacitors
Moving a charge in a field takes or gives energy. Potential difference measures this per unit charge, and a capacitor stores charge and energy in a field.
📶 Potential and uniform fields
The potential difference V between two points is the work done per unit charge: V = W / q. Between parallel plates a distance d apart the field is uniform: E = V / d.
120 V across plates 0.02 m apart
E = 120 / 0.02 = 6000 V/m
A charge of 2 μC moved through 50 V:
W = qV = 2×10⁻⁶ · 50 = 100 μJ
🖼️ See it
The time constant RC is the time to reach 63% of the supply voltage. After about 5 time constants the capacitor is effectively full.
🧊 Capacitors
A capacitor stores charge: Q = CV, where C is the capacitance in farads. The energy stored is E = ½CV².
C = 5 μF, V = 12 V → Q = 60 μC
C = 2 μF, V = 10 V → E = ½·2·100 = 100 μJ
Parallel: C = C₁ + C₂ 3 + 6 = 9 μF
Series: 1/C = 1/C₁ + 1/C₂ 3 and 6 → 1/C = 1/3 + 1/6 = 1/2 → 2 μF
Capacitors combine opposite to resistors: in parallel they add, in series use reciprocals.
🧪 Worked examples: discharge, energy and series capacitors
Discharge of a capacitor. V = V₀ e^(−t ÷ RC), where RC is the time constant τ. After one time constant the voltage has fallen to 37% of the start.
C = 100 μF, R = 10 kΩ → RC = 1.0 s; V₀ = 12 V
after 2.0 s: V = 12 × e⁻² = 12 × 0.135 = 1.6 V
time to fall to half = RC ln 2 = 0.69 s
Where does the energy go? A battery supplies QV, but the capacitor stores only ½QV. The other half is dissipated as heat in the resistance of the circuit.
C = 5.0 μF, V = 12 V
Q = CV = 60 μC
supplied = QV = 720 μJ stored = ½CV² = 360 μJ heat = 360 μJ
Series capacitors. The charge is the same on each, and the voltages divide inversely to the capacitances.
3 μF and 6 μF in series across 9 V
C = 2 μF, Q = 2 × 9 = 18 μC
V(3 μF) = Q ÷ C = 18 ÷ 3 = 6 V V(6 μF) = 3 V
Potential of a point charge. V = kQ ÷ r (a scalar), and the work to bring a charge q from infinity is q V.
🎯 Exam focus: reading graphs and explaining rates
For a discharging capacitor, the graphs of Q, V and I against t are all exponential decays with the same time constant. The gradient of the Q–t graph is the current. The area under the I–t graph is the charge.
charging through a resistor: V = V₀ (1 − e^(−t/RC))
larger R or larger C → slower charging and discharging
energy stored: ½CV² = ½QV = Q² ÷ 2C
Common mistakes. Using t = RC as the time to completely discharge (it never reaches zero). Adding capacitors in series like resistors. Forgetting the micro prefix (10⁻⁶). Believing the capacitor stores charge: the net charge is zero; it stores energy and separated charge. Using the supply energy QV as the stored energy.
Uses. Smoothing a rectified supply, timing circuits, camera flashes and backup memory.
✏️ Quick check · 1 of 11
Potential difference
Potential difference is the energy transferred per unit:
Joules per coulomb.
✏️ Quick check · 2 of 11
Capacitor energy
If the voltage across a capacitor is doubled, the stored energy becomes:
V is squared.
✍️ Type your answer · 3 of 11
A uniform field exists between plates 0.02 m apart with 120 V across them.
What is the field strength, in V/m?
Enter an exact answer, without rounding.
V/m
E = V / d.
✍️ Type your answer · 4 of 11
A 2 μC charge is moved through a potential difference of 50 V.
How much work is done, in μJ?
Enter an exact answer, without rounding.
μJ
W = qV.
✍️ Type your answer · 5 of 11
A 5 μF capacitor is charged to 12 V.
What charge does it store, in μC?
Enter an exact answer, without rounding.
μC
Q = CV.
✍️ Type your answer · 6 of 11
A 2 μF capacitor is charged to 10 V.
How much energy does it store, in μJ?
Enter an exact answer, without rounding.
μJ
E = ½CV².
✍️ Type your answer · 7 of 11
A 3 μF and a 6 μF capacitor are connected in series.
What is the equivalent capacitance, in μF?
Enter an exact answer, without rounding.
μF
Add reciprocals.
✍️ Type your answer · 8 of 11
A 100 μF capacitor discharges through a 10 kΩ resistor from an initial 12 V.
What is the voltage after 2.0 s, in volts? (e⁻² = 0.135)
Enter an exact answer, without rounding.
V
Use V = V₀ e^(−t/RC).
✍️ Type your answer · 9 of 11
A 5.0 μF capacitor is charged to 12 V from a battery through a resistor.
How much energy is dissipated as heat in the circuit, in μJ?
Enter an exact answer, without rounding.
μJ
Half the energy from the battery is lost as heat.
✍️ Type your answer · 10 of 11
A 3 μF and a 6 μF capacitor are connected in series across 9 V.
What is the potential difference across the 3 μF capacitor, in volts?
Enter an exact answer, without rounding.
V
The charge is the same on both.
✏️ Quick check · 11 of 11
Time constant
The resistor in a capacitor discharge circuit is doubled. What happens to the time constant and the initial current?
Use τ = RC and I = V/R.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
One farad equals one:
Question 2 of 5
Capacitors connected in parallel have a total capacitance that is:
Question 3 of 5
A larger plate area makes the capacitance:
Question 4 of 5
Equipotential surfaces are always:
Question 5 of 5
The electron-volt is a unit of:
Lesson 3 · Magnetism
🧭 Magnetic Fields & Forces
Moving charges create magnetic fields, and magnetic fields push on moving charges. That link powers motors and particle accelerators.
🔩 Force on a current
A wire of length L carrying current I in a field B feels a force F = B I L sin θ, where θ is the angle between the wire and the field. The direction is given by the right-hand (or Fleming left-hand) rule and is perpendicular to both.
B = 0.5 T, I = 4 A, L = 0.3 m, at 90°
F = 0.5 · 4 · 0.3 = 0.6 N
B = 0.4 T, I = 5 A, L = 0.5 m, at 30°
F = 0.4 · 5 · 0.5 · sin 30° = 0.5 N
🖼️ See it
The force is always at right angles to the motion, so a magnetic field changes a charge’s direction but does no work on it.
🌀 Moving charges and fields from currents
A charge q moving at speed v across a field B feels F = qvB at right angles to its motion, so it travels in a circle and its speed does not change. Crossed electric and magnetic fields only let through particles with v = E / B.
E = 1000 V/m, B = 0.2 T → v = 1000/0.2 = 5000 m/s
Long straight wire: B = μ₀ I / (2π r)
= 2×10⁻⁷ · I / r
I = 10 A at r = 0.05 m:
B = 2×10⁻⁷ · 10 / 0.05 = 4×10⁻⁵ T = 40 μT
Magnetic forces do no work on a moving charge because the force is always perpendicular to the velocity.
🧪 Worked examples: parallel wires, coils and charged particles
Force between parallel currents. Currents in the same direction attract. The force per unit length is F/L = μ₀ I₁ I₂ ÷ 2πd = 2 × 10⁻⁷ I₁ I₂ ÷ d.
I₁ = I₂ = 5.0 A, d = 0.10 m
F/L = 2 × 10⁻⁷ × 25 ÷ 0.10 = 5.0 × 10⁻⁵ N/m = 50 μN/m
Torque on a coil. τ = N B I A sin θ, with θ the angle between the field and the normal to the coil. It is greatest when the plane of the coil is parallel to the field.
N = 50 turns, B = 0.20 T, I = 2.0 A, A = 0.010 m², maximum torque
τ = 50 × 0.20 × 2.0 × 0.010 = 0.20 N m
Charged particles in a field. The magnetic force provides the centripetal force: q v B = m v² ÷ r, so r = m v ÷ q B.
a proton and an α particle (mass 4 m, charge 2 q) enter the same field at the same speed
r ∝ m ÷ q
r(α) ÷ r(p) = (4 ÷ 2) = 2 — the α particle follows a circle with double the radius
No work. The magnetic force is always perpendicular to the velocity, so it changes the direction but not the speed or the kinetic energy.
🎯 Exam focus: directions and applications
For direction, use Fleming’s left-hand rule for the force on a current (First finger = Field, seCond = Current, thuMb = Motion). For a negative particle, reverse the direction of the conventional current.
mass spectrometer ions pass a velocity selector (qE = qvB), then curve; r ∝ m ÷ q separates the isotopes
cyclotron the period is independent of the speed, so a fixed-frequency AC voltage accelerates the ions
Hall probe a transverse voltage across a conductor measures B
Common mistakes. Using the angle with the wrong reference: F = B I L sin θ uses the angle between the wire and the field. Forgetting that the force on a charge at rest is zero. Using F = qvB for a particle moving parallel to the field (the force is zero). Mixing up electrons (opposite to the conventional current). Writing that the magnetic force speeds up the particle.
SI unit. One tesla is the field that produces a force of 1 N on a 1 m wire carrying 1 A perpendicular to the field.
✏️ Quick check · 1 of 11
Force direction
The magnetic force on a moving charge is:
It is a cross product.
✏️ Quick check · 2 of 11
Circular path
A charged particle moving at right angles to a uniform magnetic field travels in:
The force changes direction, not speed.
✏️ Quick check · 3 of 11
Magnets
Which statement about magnetic field lines is correct?
No monopoles.
✍️ Type your answer · 4 of 11
B = 0.5 T, I = 4 A and L = 0.3 m, with the wire at right angles to the field.
What is the force, in N?
Enter an exact answer, without rounding.
N
F = BIL when θ = 90°.
✍️ Type your answer · 5 of 11
B = 0.4 T, I = 5 A, L = 0.5 m and the angle between wire and field is 30°.
What is the force, in N?
Enter an exact answer, without rounding.
N
sin 30° = 0.5.
✍️ Type your answer · 6 of 11
Crossed fields are set to E = 1000 V/m and B = 0.2 T.
What speed passes straight through, in m/s?
Enter an exact answer, without rounding.
m/s
Balance the electric and magnetic forces.
✍️ Type your answer · 7 of 11
A long straight wire carries 10 A. Use B = 2 × 10⁻⁷ · I / r.
What is B at 0.05 m, in μT?
Enter an exact answer, without rounding.
μT
1 μT = 10⁻⁶ T.
✍️ Type your answer · 8 of 11
Two long parallel wires are 0.10 m apart and each carries 5.0 A in the same direction. Use μ₀ ÷ 2π = 2 × 10⁻⁷ N A⁻².
What is the force per unit length between them, in μN/m?
Enter an exact answer, without rounding.
μN/m
Use F/L = 2 × 10⁻⁷ I₁I₂ ÷ d.
✍️ Type your answer · 9 of 11
A coil of 50 turns and area 0.010 m² carries 2.0 A in a field of 0.20 T, with its plane parallel to the field.
What is the torque on the coil, in N m?
Enter an exact answer, without rounding.
N m
Parallel to the field gives the maximum torque.
✍️ Type your answer · 10 of 11
A proton and an α particle (4 times the mass and twice the charge) move at the same speed at right angles to the same magnetic field.
The radius of the α-particle path is how many times that of the proton?
Enter an exact answer, without rounding.
Compare m/q for the two particles.
✏️ Quick check · 11 of 11
Magnetic force
A charged particle moves in a circle in a uniform magnetic field. What is true about its kinetic energy?
Think about the angle between F and v.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The SI unit of magnetic field strength is the:
Question 2 of 5
A wire parallel to a magnetic field feels a force of:
Question 3 of 5
Two parallel wires carrying current in the same direction:
Question 4 of 5
A mass spectrometer separates ions by their:
Question 5 of 5
The Earth’s magnetic field is useful for navigation because:
Lesson 4 · Magnetism
🎡 Electromagnetic Induction
A changing magnetic field makes an electric current. Induction is how power stations generate electricity and how wireless chargers work.
🧲 Faraday’s and Lenz’s laws
Magnetic flux through a loop: Φ = B A cos θ (webers). Faraday’s law: the induced emf equals the rate of change of flux linkage, ε = N ΔΦ / Δt. Lenz’s law: the induced current opposes the change that caused it (the minus sign).
B = 0.5 T, A = 0.04 m², face-on
Φ = 0.5 · 0.04 = 0.02 Wb
N = 100 turns, flux changes by 0.02 Wb in 0.5 s
ε = 100 · 0.02 / 0.5 = 4 V
🖼️ See it
A coil turning in a magnetic field cuts the flux, so an emf is induced that rises, falls and reverses: alternating current. Faster rotation raises both the peak emf and the frequency.
🚃 Moving conductors
A rod of length L moving at speed v across a field B has an emf ε = B L v. A generator turns a coil in a field, so the flux keeps changing and an alternating emf results.
B = 0.5 T, L = 0.4 m, v = 10 m/s
ε = 0.5 · 0.4 · 10 = 2 V
If the circuit resistance is 4 Ω
I = 2 / 4 = 0.5 A
Lenz’s law is conservation of energy: if the induced current helped the change instead of opposing it, energy would be created from nothing.
🧪 Worked examples: generators, motors and inductors
Rotating coil. The emf varies sinusoidally with peak ε₀ = N B A ω.
N = 100, B = 0.50 T, A = 0.040 m², ω = 100 rad/s
ε₀ = 100 × 0.50 × 0.040 × 100 = 200 V
ε = ε₀ sin ωt; the emf is a maximum when the plane of the coil is parallel to the field
(the flux is zero but changing fastest)
Back emf in a motor. A spinning armature generates an emf that opposes the supply. It limits the current.
supply 12 V, armature resistance 2.0 Ω, running current 2.0 A
back emf = 12 − I R = 12 − 2.0 × 2.0 = 8 V
at start-up the back emf is zero → the current is 12 ÷ 2.0 = 6 A (the largest)
Self-inductance. A changing current in a coil induces an emf in itself: ε = −L ΔI ÷ Δt.
L = 0.50 H, the current falls from 4.0 A to 0 in 0.10 s
|ε| = 0.50 × 4.0 ÷ 0.10 = 20 V (a large spark when a switch is opened)
Lenz’s law in action. A magnet dropped through a copper tube falls slowly because the induced eddy currents create a field that opposes the motion.
🎯 Exam focus: explaining induced emf
A full explanation uses three links: (1) the magnetic flux linkage through the coil changes; (2) Faraday’s law: an emf is induced, equal to the rate of change of flux linkage; (3) Lenz’s law: the induced current produces a field that opposes the change.
flux Φ = B A cos θ flux linkage = N Φ
ε = −N ΔΦ ÷ Δt (the minus sign is Lenz’s law)
moving rod: ε = B L v
Common mistakes. Saying that a magnet inside a coil induces an emf (it needs a changing flux). Forgetting to multiply by the number of turns. Saying that Lenz’s law “stops” the change (it opposes it). Confusing flux (Wb) with flux density (T). Believing the emf of a rotating coil is largest when the flux is largest.
Energy conservation. If the induced current helped the change, energy would be created from nothing. The opposing force means that mechanical work is converted to electrical energy.
✏️ Quick check · 1 of 11
Lenz
Lenz’s law says the induced current:
The minus sign.
✏️ Quick check · 2 of 11
Induction needs
An emf is induced in a coil when:
Change is the key.
✍️ Type your answer · 3 of 11
B = 0.5 T and a loop of area 0.04 m² face-on to the field.
What is the flux, in Wb, as a decimal?
Enter an exact answer, without rounding.
Wb
Φ = BA cos 0°.
✍️ Type your answer · 4 of 11
A coil of 100 turns has its flux change by 0.02 Wb in 0.5 s.
What is the induced emf, in V?
Enter an exact answer, without rounding.
V
Multiply by the number of turns.
✍️ Type your answer · 5 of 11
A rod 0.4 m long moves at 10 m/s across a 0.5 T field.
What emf is induced, in V?
Enter an exact answer, without rounding.
V
ε = BLv.
✍️ Type your answer · 6 of 11
The rod from the question above is part of a circuit of resistance 4 Ω and emf 2 V.
What current flows, in A?
Enter an exact answer, without rounding.
A
Ohm’s law.
✍️ Type your answer · 7 of 11
The flux through a 50-turn coil changes by 0.6 Wb in 0.2 s.
What is the emf, in V?
Enter an exact answer, without rounding.
V
ε = NΔΦ/Δt.
✍️ Type your answer · 8 of 11
A coil of 100 turns and area 0.040 m² rotates at an angular speed of 100 rad/s in a uniform field of 0.50 T.
What is the peak emf, in volts?
Enter an exact answer, without rounding.
V
Use ε₀ = NBAω.
✍️ Type your answer · 9 of 11
A motor on a 12 V supply has an armature resistance of 2.0 Ω and draws 2.0 A when running.
What is the back emf, in volts?
Enter an exact answer, without rounding.
V
Supply = back emf + I R.
✍️ Type your answer · 10 of 11
An inductor of 0.50 H has a current that falls steadily from 4.0 A to zero in 0.10 s.
What is the size of the induced emf, in volts?
Enter an exact answer, without rounding.
V
Use ε = L ΔI/Δt.
✏️ Quick check · 11 of 11
Lenz’s law
A magnet is dropped through a long vertical copper tube. Compared with free fall, it falls:
Use Lenz’s law.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The SI unit of magnetic flux is the:
Question 2 of 5
Doubling the number of turns on a coil (same flux change) makes the emf:
Question 3 of 5
A magnet is pushed into a coil. The induced current creates a field that:
Question 4 of 5
A generator converts:
Question 5 of 5
Eddy currents are:
Lesson 5 · Electricity
〰️ AC Circuits & Transformers
Mains electricity alternates in direction many times a second. Transformers change the voltage so power can travel long distances with little loss.
🌊 Alternating current
An AC voltage varies like a sine wave: V = V₀ sin(ωt). The rms value is the DC equivalent for heating: Vrms = V₀ / √2. The frequency is f = 1 / T.
T = 0.02 s → f = 1/0.02 = 50 Hz
V₀ = 10 V, ωt = 30° → V = 10 sin 30° = 5 V
P = V_rms I_rms
120 V and 2 A → 240 W
P = V_rms² / R
12 V across 24 Ω → 144/24 = 6 W
🖼️ See it
Vₛ/Vₚ = Nₛ/Nₚ. The secondary voltage is lower, so the current is higher, and in an ideal transformer the power is the same on both sides.
🔌 Transformers
A transformer uses changing flux in an iron core to change an AC voltage. For an ideal transformer: Vs / Vp = Ns / Np and power in = power out, so VpIp = VsIs.
Vp = 230 V, Np = 1000, Ns = 50
Vs = 230 · 50/1000 = 11.5 V
Primary current 0.2 A: power = 230 · 0.2 = 46 W
Is = 46 / 11.5 = 4 A
Step-up raises voltage and lowers current. Power lines use high voltage so the current, and so the I²R heating loss, is small.
🧪 Worked examples: peak values, line losses and mean power
rms and peak. V₀ = √2 × V_rms ≈ 1.41 × V_rms, and the same for current.
mains 230 V rms → peak = 230 × 1.414 = 325 V (peak-to-peak 650 V)
Mean power in a resistor. P = ½ I₀² R = I_rms² R.
I₀ = 2.0 A, R = 10 Ω → P = ½ × 4.0 × 10 = 20 W
I_rms = 2.0 ÷ 1.414 = 1.41 A; check: 1.41² × 10 = 20 W
Why transmit at high voltage. Send 10 kW through lines of total resistance 2.0 Ω.
at 200 V: I = 10 000 ÷ 200 = 50 A power lost = I² R = 2500 × 2 = 5000 W (half the power!)
at 20 000 V: I = 10 000 ÷ 20 000 = 0.50 A power lost = 0.25 × 2 = 0.50 W
Transformer efficiency = power out ÷ power in. 100 W in and 90 W out gives 90%.
Step up, step down. Step-up transformers raise the voltage and lower the current at the power station, and step-down transformers make it safe for homes.
🎯 Exam focus: losses in transformers and rectification
losses in a real transformer
• resistance of the coils → heat (I² R) use thick copper wire
• eddy currents in the core → heat laminate the core
• flux leakage wind the secondary on top of the primary
• hysteresis (magnetising and demagnetising) use a soft iron core
Rectification: a diode gives half-wave rectification. A bridge of four diodes gives full-wave. A smoothing capacitor then fills the gaps; a larger C or R gives less ripple.
Common mistakes. Using the peak value in P = V I without the factor ½. Saying that a transformer works with DC (it needs a changing flux). Believing a step-up transformer creates energy: the current falls as the voltage rises. Writing that the rms value is the average of the alternating voltage (the average over a cycle is zero).
Reading an oscilloscope. Voltage = number of divisions × volts per division; period = divisions × time base; f = 1 ÷ T.
✏️ Quick check · 1 of 11
Why AC
Power is sent along transmission lines at high voltage because:
Think I²R.
✏️ Quick check · 2 of 11
DC from a transformer
A transformer does not work with steady DC because:
Faraday’s law.
✍️ Type your answer · 3 of 11
An AC supply has a period of 0.02 s.
What is its frequency, in Hz?
Enter an exact answer, without rounding.
Hz
f = 1 / T.
✍️ Type your answer · 4 of 11
A voltage has a peak value of 10 V and V = V₀ sin(ωt).
What is V when ωt = 30°, in V?
Enter an exact answer, without rounding.
V
sin 30° = 0.5.
✍️ Type your answer · 5 of 11
A heater runs at 120 V rms and draws 2 A rms.
What is the power, in W?
Enter an exact answer, without rounding.
W
P = V_rms · I_rms.
✍️ Type your answer · 6 of 11
A transformer has 1000 primary turns and 50 secondary turns. The primary voltage is 230 V.
What is the secondary voltage, in V?
Enter an exact answer, without rounding.
V
Vs/Vp = Ns/Np.
✍️ Type your answer · 7 of 11
An ideal transformer takes 0.2 A at 230 V and delivers 11.5 V.
What current is in the secondary, in A?
Enter an exact answer, without rounding.
A
Power in = power out.
✍️ Type your answer · 8 of 11
The mains supply is 230 V rms.
What is the peak voltage, to the nearest volt?
Round your answer as the question asks.
V
V₀ = √2 × V_rms.
✍️ Type your answer · 9 of 11
A 10 kW supply is sent through cables of total resistance 2.0 Ω at 200 V.
What power is lost in the cables, in watts?
Enter an exact answer, without rounding.
W
Find the current first.
✍️ Type your answer · 10 of 11
An alternating current has a peak value of 2.0 A and flows through a 10 Ω resistor.
What is the mean power in the resistor, in watts?
Enter an exact answer, without rounding.
W
The mean power is half the peak power.
✏️ Quick check · 11 of 11
Laminated core
Why is the iron core of a transformer made from thin laminated sheets?
Think about the induced currents in the core itself.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The rms value of an AC supply is:
Question 2 of 5
A step-down transformer has:
Question 3 of 5
The laminated iron core of a transformer reduces:
Question 4 of 5
Mains frequency in many countries is:
Question 5 of 5
A real transformer is not 100% efficient because of:
Lesson 6 · Waves
🔭 Optics: Refraction, Lenses & Interference
Light bends when it enters a new material, focuses through lenses, and spreads out into patterns when it passes through narrow gaps.
🔍 Refraction and lenses
Snell’s law: n₁ sin θ₁ = n₂ sin θ₂, where n = c / v is the refractive index. Light bending away from the normal can be totally reflected beyond the critical angle: sin C = 1 / n.
air → glass n = 1.5, sin θ₁ = 0.75
sin θ₂ = 0.75 / 1.5 = 0.5 → θ₂ = 30°
n = 2: sin C = 1/2 → C = 30°
speed in glass v = c / n = 300000 / 1.5 = 200000 km/s
Thin lens: 1/f = 1/u + 1/v
f = 10 cm, u = 30 cm → 1/v = 1/10 − 1/30 = 1/15
v = 15 cm, magnification m = v/u = 0.5
Power P = 1/f (f in metres): f = 0.25 m → 4 D
🖼️ See it
Interference patterns depend on how these waves overlap: crest on crest adds, crest on trough cancels.
🌈 Interference and diffraction
In Young’s double slit experiment bright fringes are spaced x = λD / a (slit separation a, screen distance D). A diffraction grating gives bright lines where d sin θ = nλ.
λ = 600 nm, D = 2 m, a = 0.4 mm
x = 600×10⁻⁹ · 2 / 0.4×10⁻³ = 3×10⁻³ m = 3 mm
500 lines/mm → d = 2 μm = 2000 nm
λ = 500 nm, n = 2:
sin θ = 2 · 500 / 2000 = 0.5 → θ = 30°
Constructive interference occurs when the path difference is a whole number of wavelengths, and destructive when it is a half-odd number.
🧪 Worked examples: lenses, critical angles and diffraction
Thin lens (real-is-positive). 1/f = 1/u + 1/v, and the magnification is m = v ÷ u.
converging lens f = 10 cm, object at u = 15 cm
1/v = 1/10 − 1/15 = 1/30 → v = 30 cm (real, inverted)
m = 30 ÷ 15 = 2 (twice as large)
Critical angle. sin C = 1/n.
n = 1.25 → sin C = 0.80 → C = 53°
light inside the material hitting the surface at more than 53° is totally reflected
Young’s fringes. x = λD ÷ a. Doubling the slit separation halves the fringe spacing: 3.0 mm becomes 1.5 mm.
Diffraction grating. d sin θ = nλ. The highest order is the largest whole n with nλ ≤ d.
d = 2000 nm, λ = 500 nm: n ≤ 2000 ÷ 500 = 4
n = 4 would be at θ = 90° and cannot be seen, so the highest visible order is 3
number of bright lines in total = 2 × 3 + 1 = 7
Sign convention. Real images have positive v. A diverging lens has a negative f and always gives a virtual, upright, smaller image.
🎯 Exam focus: coherence, polarisation and explanations
For clear interference fringes the sources must be coherent: the same frequency and a constant phase difference. Two separate lamps are not coherent; two slits lit by the same source are.
polarisation: Malus’s law I = I₀ cos² θ (60° → I = I₀ × 0.25)
only transverse waves can be polarised: this shows that light is a transverse wave
single-slit diffraction: the central maximum is wider if the slit is narrower
Common mistakes. Forgetting to convert nm and mm into metres. Using the wrong sign for v or f in the lens equation. Confusing the angle in Snell’s law with the angle to the surface (always use the normal). Saying that the fringe spacing increases when the wavelength decreases (it decreases). Including the order n = 4 when sin θ would be 1.
Path difference. Bright fringes occur when the path difference is nλ, dark ones at (n + ½)λ.
✏️ Quick check · 1 of 11
Total internal reflection
Total internal reflection can occur only when light travels:
Think about the critical angle.
✏️ Quick check · 2 of 11
Interference
Two waves meet in phase. The result is:
In phase means crest to crest.
✍️ Type your answer · 3 of 11
Light goes from air into glass of n = 1.5 with sin θ₁ = 0.75.
What is sin θ₂, as a decimal?
Enter an exact answer, without rounding.
Snell’s law.
✍️ Type your answer · 4 of 11
The refractive index of a material is 2.
What is the critical angle, in degrees?
Enter an exact answer, without rounding.
°
sin C = 1/n.
✍️ Type your answer · 5 of 11
A converging lens has f = 10 cm and the object is 30 cm away.
Where is the image, in cm from the lens?
Enter an exact answer, without rounding.
cm
1/f = 1/u + 1/v.
✍️ Type your answer · 6 of 11
The lens in the question above has v = 15 cm and u = 30 cm.
What is the magnification?
Enter an exact answer, without rounding.
m = v / u.
✍️ Type your answer · 7 of 11
λ = 600 nm, D = 2 m and a = 0.4 mm in a double-slit experiment.
What is the fringe spacing, in mm?
Enter an exact answer, without rounding.
mm
Convert everything to metres.
✍️ Type your answer · 8 of 11
A converging lens has a focal length of 10 cm. An object is placed 15 cm from it.
How far from the lens is the image, in cm?
Enter an exact answer, without rounding.
cm
Use 1/f = 1/u + 1/v.
✍️ Type your answer · 9 of 11
A transparent material has a refractive index of 1.25.
What is the critical angle for light going from the material to air, to the nearest degree?
Round your answer as the question asks.
°
sin C = 1/n.
✍️ Type your answer · 10 of 11
A grating has a slit spacing of 2000 nm and is illuminated with light of wavelength 500 nm.
What is the highest order of bright line that can be observed?
Enter an exact answer, without rounding.
Use d sin θ = nλ with sin θ < 1.
✏️ Quick check · 11 of 11
Polarisation
What does the polarisation of light show about its nature?
Which type of wave can be filtered by a slit-like polariser?
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
The refractive index of a material is:
Question 2 of 5
A lens with a short focal length is:
Question 3 of 5
Light diffracts most when the gap is:
Question 4 of 5
Which colour of visible light has the longest wavelength?
Question 5 of 5
Polarisation shows that light is a:
Lesson 7 · Modern physics
💡 Quantum Physics
Light and matter behave as both waves and particles. Quantum ideas explain the photoelectric effect and the colours of atoms.
🔦 Photons and the photoelectric effect
Light comes in packets called photons with energy E = hf, or in convenient units E (eV) = 1240 / λ (nm). A photon can eject an electron from a metal only if its energy exceeds the work function φ.
λ = 620 nm → E = 1240/620 = 2 eV
KE_max = E − φ
photon 5 eV, φ = 2 eV → KE_max = 3 eV
stopping potential = 3 V
threshold wavelength for φ = 2.5 eV
λ = 1240 / 2.5 = 496 nm
Brighter light releases more electrons per second but not faster ones. Only a higher frequency increases the maximum kinetic energy.
🖼️ See it
Below the threshold frequency no electrons are released, however bright the light. Above it, energy rises linearly with frequency.
⚛️ Energy levels of hydrogen
Electrons in atoms have only certain energies. In hydrogen Eₙ = −13.6 / n² eV. A photon is emitted when an electron falls to a lower level, with energy equal to the difference.
n = 1: −13.6 eV n = 2: −3.4 eV
ionisation energy from the ground state = 13.6 eV
transition n = 2 → 1
E = −3.4 − (−13.6) = 10.2 eV
λ = 1240 / 10.2 ≈ 122 nm (ultraviolet)
🧪 Worked examples: thresholds, matter waves and spectra
Threshold wavelength. The photon energy must at least equal the work function. With E (eV) = 1240 ÷ λ (nm):
φ = 2.0 eV → λ₀ = 1240 ÷ 2.0 = 620 nm
light of 700 nm (1.77 eV) ejects nothing, however bright;
light of 400 nm (3.10 eV) gives KE_max = 3.10 − 2.0 = 1.10 eV
de Broglie wavelength. λ = h ÷ p. Matter behaves as a wave.
an electron with p = 3.3 × 10⁻²⁴ kg m/s
λ = 6.6 × 10⁻³⁴ ÷ 3.3 × 10⁻²⁴ = 2.0 × 10⁻¹⁰ m = 0.20 nm (comparable to atomic spacing → electron diffraction)
Hydrogen emission lines.
n = 3 → n = 2: ΔE = 13.6 × (1/4 − 1/9) = 13.6 × 5/36 = 1.89 eV
λ = 1240 ÷ 1.89 = 656 nm (the red Balmer line)
Photon flux. A 3.2 mW beam of 2.0 eV photons (3.2 × 10⁻¹⁹ J each): N = 3.2 × 10⁻³ ÷ 3.2 × 10⁻¹⁹ = 1.0 × 10¹⁶ photons per second.
Stopping potential. The electron’s maximum kinetic energy, in eV, equals the stopping voltage in volts. A graph of KE_max against frequency has gradient h and an intercept −φ.
🎯 Exam focus: why the wave theory fails
The photoelectric effect gives three observations that classical wave theory cannot explain:
1 a threshold frequency exists: below it, no emission, whatever the intensity
2 KE_max depends on the frequency, not on the intensity
3 emission is instant, with no time delay for energy to build up
explanation: each electron absorbs one photon of energy hf; if hf < φ, no emission
Common mistakes. Saying that a brighter beam gives faster electrons (it gives more electrons per second). Confusing work function and ionisation energy. Using the wavelength in nm in E = hc ÷ λ without conversion. Forgetting that the stopping potential is for the fastest electrons. Saying that electrons are “either waves or particles”: they show both behaviours depending on the experiment.
Line spectra are evidence for discrete energy levels in atoms. A continuous spectrum comes from a hot dense solid, an absorption spectrum has dark lines at the same wavelengths as the emission lines.
✏️ Quick check · 1 of 11
Photon energy
The energy of a photon is proportional to its:
E = hf.
✏️ Quick check · 2 of 11
Threshold
No electrons are emitted from a metal when the photon energy is:
Photon energy versus φ.
✏️ Quick check · 3 of 11
Wave–particle
Electrons can show diffraction patterns, which shows that they have:
De Broglie.
✍️ Type your answer · 4 of 11
Light has a wavelength of 620 nm. Use E = 1240 / λ.
What is the photon energy, in eV?
Enter an exact answer, without rounding.
eV
E (eV) = 1240 / λ (nm).
✍️ Type your answer · 5 of 11
A photon of 5 eV hits a metal with work function 2 eV.
What is the maximum kinetic energy of an emitted electron, in eV?
Enter an exact answer, without rounding.
eV
KE = E − φ.
✍️ Type your answer · 6 of 11
Electrons are emitted with a maximum kinetic energy of 3 eV.
What stopping potential is needed, in V?
Enter an exact answer, without rounding.
V
eV = e × volts.
✍️ Type your answer · 7 of 11
A hydrogen electron falls from n = 2 to n = 1 (Eₙ = −13.6/n² eV).
What is the energy of the photon, in eV?
Enter an exact answer, without rounding.
eV
Subtract the two energies.
✍️ Type your answer · 8 of 11
A metal has a work function of 2.0 eV. Use E (eV) = 1240 ÷ λ (nm).
What is the threshold wavelength, in nm?
Enter an exact answer, without rounding.
nm
The photon energy equals the work function.
✍️ Type your answer · 9 of 11
An electron has a momentum of 3.3 × 10⁻²⁴ kg m/s. Use h = 6.6 × 10⁻³⁴ J s.
What is its de Broglie wavelength, in nm?
Enter an exact answer, without rounding.
nm
Use λ = h/p.
✍️ Type your answer · 10 of 11
An electron in a hydrogen atom drops from n = 3 to n = 2, and E_n = −13.6 ÷ n² eV.
What is the wavelength of the photon emitted, in nm, to the nearest nm?
Round your answer as the question asks.
nm
Find the energy difference first.
✏️ Quick check · 11 of 11
Photoelectric effect
The intensity of light above the threshold frequency is doubled. What happens to the photoelectrons?
Think of photons as separate packets.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Increasing the intensity of light above the threshold frequency increases:
Question 2 of 5
The work function of a metal is:
Question 3 of 5
An atom emits a line spectrum because:
Question 4 of 5
Which has the most energetic photons?
Question 5 of 5
The de Broglie wavelength of a particle gets shorter when its momentum:
Lesson 8 · Modern physics
☢️ Nuclear Physics & Relativity
The nucleus holds enormous energy, and at high speeds time and length no longer behave the way everyday experience suggests.
🧪 Radioactive decay and binding energy
Unstable nuclei decay by emitting alpha (⁴₂He), beta (an electron, so Z rises by one) or gamma (high-energy photons). Both mass number A and atomic number Z are conserved. The number of undecayed nuclei halves every half-life.
²³⁸U (Z = 92) → α → A = 238 − 4 = 234, Z = 92 − 2 = 90
640 nuclei, half-life 2 h, after 6 h (3 half-lives)
N = 640 / 2³ = 80
mass defect → energy: E = Δm c²
use 1 u = 930 MeV: Δm = 0.02 u → 18.6 MeV
🖼️ See it
Every half-life removes half of what is left, so the curve never quite reaches zero.
🚀 Special relativity
Einstein’s postulates: the laws of physics are the same for all uniformly moving observers, and the speed of light is the same for all. The result is time dilation and length contraction, with the factor γ = 1 / √(1 − v²/c²).
v = 0.6c → γ = 1/√(1 − 0.36) = 1/0.8 = 1.25
proper time 8 s → observed t = γ · 8 = 10 s
proper length 10 m → L = 10 / γ = 8 m
Proper time and proper length are measured in the frame where the clock or object is at rest. Everyone else sees time stretch and length shrink.
🧪 Worked examples: activity, binding energy and relativity
Activity halves every half-life (like the number of undecayed nuclei). A = λN, with λ = ln 2 ÷ T½.
A₀ = 1.0 × 10⁶ Bq, half-life 30 min; after 2 h = 4 half-lives
A = 1.0 × 10⁶ ÷ 2⁴ = 62 500 Bq
Binding energy from the mass defect. E = Δm c², or with 1 u = 931.5 MeV:
mass defect 0.030 u → E = 0.030 × 931.5 = 27.9 MeV
for 4 nucleons (helium-4): 27.9 ÷ 4 = 7.0 MeV per nucleon
Time dilation and length contraction at v = 0.8c: γ = 1 ÷ √(1 − 0.64) = 1 ÷ 0.6 = 1.67.
proper time 6.0 s → observed time = 1.67 × 6.0 = 10 s
proper length 10 m → observed length = 10 ÷ 1.67 = 6.0 m
Mass–energy. A mass loss of 1.0 × 10⁻³ kg releases E = m c² = 1.0 × 10⁻³ × 9.0 × 10¹⁶ = 9.0 × 10¹³ J.
Muons. Cosmic-ray muons live only 2.2 μs, but reach the ground because time dilation makes their lifetime longer in the Earth’s frame.
🎯 Exam focus: radiation properties and nuclear equations
α β⁻ γ
ionising very high medium low
penetration paper / few cm few mm aluminium thick lead / concrete
deflection small (positive) large (negative) none
balance A and Z: ¹⁴₆C → ¹⁴₇N + ⁰₋₁e + ν̄
Radioactive decay is random and spontaneous: you cannot predict when a given nucleus will decay, only the probability per unit time (the decay constant λ).
Common mistakes. Forgetting the antineutrino in β⁻ decay. Using the mass number instead of the atomic number to balance charge. Taking the half-life as half the time for all the nuclei to decay. Mixing up proper time (measured in the frame where the events occur in the same place) with observed time. Forgetting that fission and fusion both release energy because the products have a higher binding energy per nucleon.
Background radiation. Subtract the background count rate from the measured rate before using it to find the half-life.
✏️ Quick check · 1 of 11
Alpha particle
An alpha particle consists of:
It is a helium-4 nucleus.
✏️ Quick check · 2 of 11
Beta decay
In beta-minus decay the atomic number Z of the nucleus:
A neutron becomes a proton.
✏️ Quick check · 3 of 11
Speed of light
According to special relativity, the speed of light in a vacuum is:
It is invariant.
✍️ Type your answer · 4 of 11
A uranium-238 nucleus (Z = 92) emits an alpha particle.
What is the mass number of the daughter nucleus?
Enter an exact answer, without rounding.
An alpha removes 4 nucleons.
✍️ Type your answer · 5 of 11
A sample contains 640 radioactive nuclei with a half-life of 2 hours.
How many remain after 6 hours?
Enter an exact answer, without rounding.
Halve it three times.
✍️ Type your answer · 6 of 11
The mass defect of a nucleus is 0.02 u. Use 1 u = 930 MeV.
What is the binding energy, in MeV?
Enter an exact answer, without rounding.
MeV
E = Δm c², expressed in MeV.
✍️ Type your answer · 7 of 11
A spaceship moves at 0.6c, so γ = 1.25. A clock on board measures a proper time of 8 s.
How long does that take for a stationary observer, in s?
Enter an exact answer, without rounding.
s
Time is dilated: multiply by γ.
✍️ Type your answer · 8 of 11
A source has an activity of 1.0 × 10⁶ Bq and a half-life of 30 minutes.
What is its activity after 2 hours, in becquerels?
Enter an exact answer, without rounding.
Bq
Count the half-lives.
✍️ Type your answer · 9 of 11
A nucleus has a mass defect of 0.030 u. Use 1 u = 931.5 MeV.
What is its binding energy, in MeV?
Enter an exact answer, without rounding.
MeV
Multiply the mass defect by 931.5.
✍️ Type your answer · 10 of 11
A spacecraft clock records 6.0 s between two events. The craft moves at 0.8c relative to Earth.
How long between the events is measured on Earth, in seconds?
Enter an exact answer, without rounding.
s
Proper time is shorter than the observed time.
✏️ Quick check · 11 of 11
Radiation
Which type of radiation has the highest ionising power and the lowest penetration?
The largest and slowest particle.
📝 Lesson quiz · 5 questions
0 / 5
Re-read the lesson, then give it another go.
Question 1 of 5
Gamma radiation is:
Question 2 of 5
After one half-life, the remaining fraction of a radioactive sample is:
Question 3 of 5
Nuclear fission is the:
Question 4 of 5
E = mc² shows that:
Question 5 of 5
As a spaceship approaches the speed of light, γ:
Final Challenge
🏆 16-Question Challenge Quiz
Questions from every topic in this course. Pick an answer for each one, then hit Check My Answers.
0 / 16
💪 Keep practicing — go back through the lessons and retake the quiz!
Q1 · Electric Fields & Coulomb’s Law
The force between two point charges is proportional to:
Q2 · Electric Fields & Coulomb’s Law
An object becomes positively charged by:
Q3 · Electric Potential & Capacitors
The energy stored in a capacitor is:
Q4 · Electric Potential & Capacitors
The electron-volt is a unit of:
Q5 · Magnetic Fields & Forces
A moving charge in a magnetic field experiences a force only if:
Q6 · Magnetic Fields & Forces
The Earth’s magnetic field is useful for navigation because:
Q7 · Electromagnetic Induction
The emf induced in a coil is proportional to the:
Q8 · Electromagnetic Induction
Eddy currents are:
Q9 · AC Circuits & Transformers
For an ideal transformer, the ratio Vs/Vp equals:
Q10 · AC Circuits & Transformers
A real transformer is not 100% efficient because of:
Q11 · Optics: Refraction, Lenses & Interference
Total internal reflection is used in:
Q12 · Optics: Refraction, Lenses & Interference
Polarisation shows that light is a:
Q13 · Quantum Physics
Einstein explained the photoelectric effect by treating light as:
Q14 · Quantum Physics
The de Broglie wavelength of a particle gets shorter when its momentum: